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In a potentiometer experiment, the balancing length of a cell is $560 \mathrm{~cm}$. When an external resistance of $10 \Omega$ is connected in parallel to the cell, the balancing length changes by 60 $\mathrm{cm}$. The internal resistance of a cell is
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The correct answer is:
$1.2 \Omega$
Let us Consider a cell of emf E and balancing length $1_{1}$
$\mathrm{E}=\mathrm{kl}_{1}$
potential difference is balanced by length $\mathrm{l}_{2}$. $\mathrm{V}=\mathrm{kl}_{2}$
Internal resistance of the cell
$\begin{array}{l}
\mathrm{r}=\left(\frac{\mathrm{E}-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}=\left(\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}}-1\right) \mathrm{R} \\
=\left(\frac{560}{560-60}-1\right) 10=\left(\frac{56}{50}-1\right) 10 \\
=\frac{6}{5}=1.2 \Omega
\end{array}$
$\mathrm{E}=\mathrm{kl}_{1}$
potential difference is balanced by length $\mathrm{l}_{2}$. $\mathrm{V}=\mathrm{kl}_{2}$
Internal resistance of the cell
$\begin{array}{l}
\mathrm{r}=\left(\frac{\mathrm{E}-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}=\left(\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}}-1\right) \mathrm{R} \\
=\left(\frac{560}{560-60}-1\right) 10=\left(\frac{56}{50}-1\right) 10 \\
=\frac{6}{5}=1.2 \Omega
\end{array}$
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